Question

prove that
[1+ CosA-Casec A] [1+tan A+ sec A] = 2
2​

Answer 1

Correct question :

prove that

(1 + CotA - CasecA)(1 + tanA + secA) = 2

Solution

LHS = (1 + CotA - CosecA) (1 + tanA + secA)

=> 1 + tanA + secA + cotA + cotA tanA + cotA secA - cosecA - cosecA tanA - cosecA secA

= > 1 + tanA + secA + cotA + 1 + cosecA - cosecA - secA - cosecA secA

=> 2 + tanA + cotA - secA cosecA

=> 2 + (sinA/cosA) + (cosA/sinA) - secA cosecA

=> 2 + (sin²A + cos²A/cosA sinA) - 1/cosA sinA

=> 2 + (1/cosA sinA) - (1/cosA sinA)

=> 2 ......RHS

Answer 2

Step-by-step explanation:

$\bf{\underline{\underline\purple{Correct\: Question:-}}}$

Prove that:-

[ 1 + cot A - cosec A ] [1 + tan A + sec A ] = 2

Basics needed to prove the above trigonometric value:-

• $\rm \cot A = \dfrac{ \cos A}{ \sin A}$

• $\rm \tan A = \dfrac{ \sin A}{ \cos A}$

• $\rm \cosec A = \dfrac{ 1}{ \sin A}$

• $\rm \sec A = \dfrac{ 1}{ \cos A}$

• $\rm { \sin}^{2} A + { \cos}^{2} A = 1$

• $\rm (a + b)(a - b) = { a}^{2} - { b}^{2}$

$\bf{\underline{\underline\red{Proof:-}}}$

$\rm[ 1 + cot A - cosec A ] [1 + tan A + sec A ] = 2$

Let us prove by simplifying LHS

LHS = [ 1 + cot A - cosec A ] [1 + tan A + sec A ]

= $\rm \bigg [1 + \dfrac{cos A}{sinA} - \dfrac{1}{sinA} \bigg] \bigg [1 + \dfrac{sinA}{cosA} + \dfrac{1}{cosA} \bigg]$

= $\rm \bigg [ \dfrac{sinA + cos A - 1}{sinA} \bigg] \bigg [ \dfrac{sinA + cosA + 1}{cosA} \bigg]$

= $\rm \dfrac{(sinA + cos A - 1)(sinA + cosA + 1)}{sinAcosA}$

= $\rm \dfrac{(sinA + cos A)^{2} - {1}^{2} }{sinAcosA}$

= $\rm \dfrac{sinA^{2} + cos A^{2} + 2sinAcosA- {1}}{sinAcosA}$

= $\rm \dfrac{1+ 2sinAcosA- {1}}{sinAcosA}$

= $\rm \dfrac{ 2sinAcosA}{sinAcosA}$

= $\rm 2 = LHS$

LHS = RHS

[ 1 + cot A - cosec A ] [1 + tan A + sec A ] = 2

Hence; Proved